Question

Find the point on x- axis which are at distance of 2sqrt 5 from point ( 7,-4) . How many such pointe are there

Answer 1

Co-ordinates of point on x-axis be (x, 0) which is at a distance of 2√5 from point (7, –4).

Distance between two points A(x1, y1) and B(x2, y2) is AB = ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2

It is given that the distance between points (x, 0) and (7, –4) is 2√5.

⇒ [ ( 7 -x )2 + ( -4 – 0)2 ] = 2√5 .

Squaring both the sides, we obtain

⇒ ( 7 – x)2 + 16 = 20

⇒ 49 + x2 – 14x + 16 = 20

⇒ x2 – 14x + 45 = 0

⇒ x2 – 5x – 9x + 45 = 0

⇒ x( x – 5) – 9( x – 5) = 0

⇒ ( x – 9) ( x – 5) = 0

∴ x = 5 and 9.

Hence, there are two points i.e., (5, 0) and (9, 0).

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